Adding fractions in mixed number form Carefully read the following methods, and make sure you understand the examples: ο· Method 1: Convert all mixed numbers to top-heavy (improper) fractions, then use your usual method for adding (find a common denominator, then add numerators). ο· Method 2: Write each mixed number separately as whole number plus fraction, then add the whole numbers separately, add the fractions separately then combine. Eg: 2 7 3 +4 5 10 Using method 1: Converting to top-heavy fractions: 2 17 7 47 3 = πππ 4 = 5 5 10 10 Finding a common denominator and adding: 17 47 34 47 ππ + = + = 5 10 10 10 ππ Using method 2: Splitting whole number and fraction parts: 2 7 2 7 3 +4 =3+ +4+ 5 10 5 10 Adding the whole numbers and the fractions separately: 2 7 4 7 11 1 3 + 4 = 7 πππ + = + = =1+ 5 10 10 10 10 10 Adding the two parts back together at the end: 1 1 π 7+1+ =8+ =π 10 10 ππ 81 1 Note that and 8 are just different ways of writing the same number (Itβs also 8.1). 10 10 1. Use method 1 to answer the following questions. You may leave your answer as a top-heavy fraction, but you should simplify as far as possible. 2 3 3 +1 = 25 10 3 2 2 +3 = 4 5 2. Use method 2 to answer the following questions. You should leave your answer in mixed number form. The fraction part should be fully simplified. 1 2 6 +7 = 4 7 5 4 20 + 34 = 6 9 Mixed Numbers Mixed Numbers You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 is not a sensible form, so convert to 6 . You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 is not a sensible form, so convert to 6 . Challenge: Challenge: 3 3 1 3 4 β2 = 3 4 3 3 1 3 4 β2 = 3 4 Mixed Numbers Mixed Numbers You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 3 is not a sensible form, so convert to 6 3. You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 3 is not a sensible form, so convert to 6 3. Challenge: Challenge: 1 3 4 β2 = 3 4 1 3 4 β2 = 3 4 Adding fractions in mixed number form SOLUTIONS Carefully read the following methods, and make sure you understand the examples: ο· Method 1: Convert all mixed numbers to top-heavy (improper) fractions, then use your usual method for adding (find a common denominator, then add numerators). ο· Method 2: Write each mixed number separately as whole number plus fraction, then add the whole numbers separately, add the fractions separately then combine. Eg: 2 7 3 +4 5 10 Using method 1: Converting to top-heavy fractions: 2 17 7 47 3 = πππ 4 = 5 5 10 10 Finding a common denominator and adding: 17 47 34 47 ππ + = + = 5 10 10 10 ππ Using method 2: Splitting whole number and fraction parts: 2 7 2 7 3 +4 =3+ +4+ 5 10 5 10 Adding the whole numbers and the fractions separately: 2 7 4 7 11 1 3 + 4 = 7 πππ + = + = =1+ 5 10 10 10 10 10 Adding the two parts back together at the end: 1 1 π 7+1+ =8+ =π 10 10 ππ 81 1 Note that and 8 are just different ways of writing the same number (Itβs also 8.1). 10 10 1. Use method 1 to answer the following questions. You may leave your answer as a top-heavy fraction, but you should simplify as far as possible. 2 3 3 53 13 212 130 342 πππ +1 = + = + = = 25 10 25 10 100 100 100 ππ 3 2 11 17 55 68 πππ 2 +3 = + = + = 4 5 4 5 20 20 ππ 2. Use method 2 to answer the following questions. You should leave your answer in mixed number form. The fraction part should be fully simplified. 1 2 1 2 7 8 15 ππ 6 + 7 = (6 + 7) + ( + ) = (13) + ( + ) = (13) + ( ) = ππ 4 7 4 7 28 28 28 ππ 5 4 5 4 15 12 27 20 + 34 = (20 + 34) + ( + ) = 54 + ( + ) = 54 + ( ) 6 9 6 9 18 18 18 3 1 π = 54 + ( ) = 54 + (1 + ) = ππ 2 2 π Mixed Numbers SOLUTIONS Mixed Numbers SOLUTIONS You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 is not a sensible form, so convert to 6 . 3 3 3 6 15 3 ππ 1) 5 25 + 6 10 = 11 + 50 + 50 = ππ ππ 3 2 15 8 3 23 3 2 3 10 21 π 3 7 3 π 6 15 3 ππ 2 15 8 3 23 π 10 3 2 3 10 21 3 7 π 3) 1 30 + 5 6 = 6 + 6 + 6 = π π 41 ππ 4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = π ππ 6 3 2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = π ππ 3) 1 30 + 5 6 = 6 + 6 + 6 = π π 20 3 3 1) 5 25 + 6 10 = 11 + 50 + 50 = ππ ππ 2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = π ππ 10 You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 is not a sensible form, so convert to 6 . 7 13 20 41 ππ 4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = π ππ π 5) 1 6 + 9 12 = 10 + 12 + 12 = 10 + 12 = ππ ππ Challenge: 1 3 4 β2 = 3 4 1 3 1 3 (4 + ) β (2 + ) = 2 + ( β ) 3 4 3 4 4 9 5 π =2+( β )=2β =π 12 12 12 ππ 6 7 13 π 5) 1 6 + 9 12 = 10 + 12 + 12 = 10 + 12 = ππ ππ Challenge: 1 3 4 β2 = 3 4 1 3 1 3 (4 + ) β (2 + ) = 2 + ( β ) 3 4 3 4 4 9 5 π = 2+( β )=2β =π 12 12 12 ππ Mixed Numbers SOLUTIONS Mixed Numbers SOLUTIONS You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 3 is not a sensible form, so convert to 6 3. You can use any method you like to answer these. If the fraction part of your answer is top-heavy, you will need to turn it into a mixed number. 4 1 Eg: 5 3 is not a sensible form, so convert to 6 3. 1) 5 25 + 6 10 = 11 + 50 + 50 = ππ ππ 1) 5 25 + 6 10 = 11 + 50 + 50 = ππ ππ 3 3 3 2 6 15 15 8 3 ππ 23 π 3 2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = π ππ 10 3 2 3 10 21 3 7 π 7 2 6 15 15 8 3 ππ 23 π 10 3 2 3 10 21 3 7 π 3) 1 30 + 5 6 = 6 + 6 + 6 = π π 41 ππ 4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = π ππ 6 3 2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = π ππ 3) 1 30 + 5 6 = 6 + 6 + 6 = π π 20 3 13 20 41 ππ 4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = π ππ π 5) 1 + 9 = 10 + + = 10 + = ππ 6 12 12 12 12 ππ Challenge: 1 3 4 β2 = 3 4 1 3 1 3 (4 + ) β (2 + ) = 2 + ( β ) 3 4 3 4 4 9 5 π =2+( β )=2β =π 12 12 12 ππ 6 7 13 π 5) 1 + 9 = 10 + + = 10 + = ππ 6 12 12 12 12 ππ Challenge: 1 3 4 β2 = 3 4 1 3 1 3 (4 + ) β (2 + ) = 2 + ( β ) 3 4 3 4 4 9 5 π = 2+( β )=2β =π 12 12 12 ππ
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